Physics 272: Advanced Honors Physics II
due in class Monday, February 12, 2018
Reading: Purcell 3.1-7
the solution for each problem and your name should be
written on a separate page (sorry, trees!) to facilitate
the grading. Please paperclip the pages together.
I have put the answers at the bottom of this page. You
can use them to check your own answers -- if they don't
agree, you know you need to go back and look for at
least one mistake.
1. A metal sphere of radius R, carrying charge q, is
surrounded by a thick concentric metal shell (inner
radius a, outer radius b, as in the left figure above).
The shell carries no net charge.
(a) Find the surface charge density sigma at R, at a,
and at b.
(b) Find the potential at the center, using infinity as
the reference point.
(c) Now the outer surface is touched to a grounding
wire, which drains off charge and lowers the potential
to zero (same as at infinity). How do your answers to
(a) and (b) change?
2. Two spherical cavities, of radii a and b, are
hollowed out from the interior of a (neutral) conducting
sphere of radius R (as in the right figure above). At
the center of each cavity a point charge is placed --
call these charges qa and qb.
(a) Find the surface charge densities sigma-a, sigma-b
(b) What is the field outside the conductor?
(c) What is the field within each cavity?
(d) What is the force on qa and qb?
(e) Which of these answers would change if a third
charge qc were brought near the conductor?
3. **3.54 Dividing
the surface charge
4. **3.60 A three-shell capacitor
5. **3.63 Capacitance coefficients for shells
6. *3.38 Two charges and a plane
7. **3.43 Images from three planes
8. **3.68 Maximum energy storage between cylinders
Note that in some cases these are partial answers --
just enough for you to check that you are doing the
1. (a) At a, the surface charge density is -q/(4 pi a2);
(c) the surface charge density at a is unchanged
2. (b) E has magnitude k|qa+qb|/r2, radially
outward if qa+qb>0.
3. sigma1 = (8/13) sigma
4. Q1 = Q/4
5. C11 = 4 pi epsilon0 ab/(a-b); C12 = -4
pi epsilon0 ab/(a-b)
6. y = (0.306)l
7. 0.210 Q2 /(4 pi epsilon0 d2)
8. U=pi epsilon0 a2 E2 / (2e) where e is
the base of the natural logarithm.